$g(n) = -7n^{3}+4n^{2}-n+2(f(n))$ $f(x) = 5x-3+3(h(x))$ $h(n) = -2n^{2}$ $ g(f(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = (5)(0)-3+3(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = -2(0^{2})$ $h(0) = 0$ That means $f(0) = (5)(0)-3+(3)(0)$ $f(0) = -3$ Now we know that $f(0) = -3$ . Let's solve for $g(f(0))$ , which is $g(-3)$ $g(-3) = -7(-3)^{3}+4(-3)^{2}-(-3)+2(f(-3))$ To solve for the value of $g$ , we need to solve for the value of $f(-3)$ $f(-3) = (5)(-3)-3+3(h(-3))$ To solve for the value of $f$ , we need to solve for the value of $h(-3)$ $h(-3) = -2(-3)^{2}$ $h(-3) = -18$ That means $f(-3) = (5)(-3)-3+(3)(-18)$ $f(-3) = -72$ That means $g(-3) = -7(-3)^{3}+4(-3)^{2}-(-3)+(2)(-72)$ $g(-3) = 84$